1. Quadratic Equations
Quadratic Equations
MODULE - I
Algebra
2
Notes
QUADRATIC EQUATIONS
Recall that an algebraic equation of the second degree is written in general form as
ax 2 + bx + c = 0, a ≠ 0
It is called a quadratic equation in x. The coefficient ‘a’ is the first or leading coefficient, ‘b’
is the second or middle coefficient and ‘c’ is the constant term (or third coefficient).
5 1
For example, 7x² + 2x + 5 = 0, x² + x + 1 = 0,
2 2
1
3x² − x = 0, x² + = 0, 2 x² + 7x = 0, are all quadratic equations.
2
In this lesson we will discuss how to solve quadratic equations with real and complex
coefficients and establish relation between roots and coefficients. We will also find cube
roots of unity and use these in solving problems.
OBJECTIVES
After studying this lesson, you will be able to:
• solve a quadratic equation with real coefficients by factorization and by using quadratic
formula;
• find relationship between roots and coefficients;
• form a quadratic equation when roots are given; and
• find cube roots of unity.
EXPECTED BACKGROUND KNOWLEDGE
• Real numbers
• Quadratic Equations with real coefficients.
MATHEMATICS 39

2. Quadratic Equations
MODULE - I
2.1 ROOTS OF A QUADRATIC EQUATION
Algebra
The value which when substituted for the variable in an equation, satisfies it, is called a root
(or solution) of the equation.
If  be one of the roots of the quadratic equation
Notes then, ax 2 + bx + c = 0, a ≠ 0 ... (i)
a 2 + b + c = 0
In other words, x −  is a factor of the quadratic equation (i)
In particular, consider a quadratic equation x2 + x − 6 = 0 ...(ii)
If we substitute x = 2 in (ii), we get
L.H.S = 22 + 2 – 6 = 0
∴ L.H.S = R.H.S.
Again put x = − 3 in (ii), we get
L.H.S. = ( − 3)2 –3 –6 = 0
∴ L.H.S = R.H.S.
Again put x = − 1 in (ii) ,we get
L.H.S = ( − 1)2 + ( − 1) – 6 = –6 ≠ 0 = R.H.S.
∴ x = 2 and x = − 3 are the only values of x which satisfy the quadratic equation (ii)
There are no other values which satisfy (ii)
∴ x = 2, x = − 3 are the only two roots of the quadratic equation (ii)
Note: If  ,  be two roots of the quadratic equation
ax 2 + bx + c = 0, a ≠ 0 ...(A)
then (x −  ) and (x −  ) will be the factors of (A). The given quadratic
equation can be written in terms of these factors as (x −  ) (x −  ) = 0
2.2 SOLVING QUADRATIC EQUATION BY FACTORIZATION
Recall that you have learnt how to factorize quadratic polynomial of the form
p ( x ) = ax 2 + bx + c , a ≠ 0, by splitting the middle term and taking the common factors.
Same method can be applied while solving quadratic equation by factorization.
p r
If x − q and x − s are two factors of the quadratic equation
p r
ax2 + bx + c = 0 ,a ≠ 0 then ( x − q )( x − s ) = 0
p r
∴ either x = q or, x = s
40 MATHEMATICS

3. Quadratic Equations
Quadratic Equations
MODULE - I
p r
∴ 2
The roots of the quadratic equation ax + bx + c = 0 are q , s Algebra
Example 2.1 Using factorization method, solve the following quadratic equation :
6x 2 + 5x − 6 = 0
Solution: The given quadratic equation is Notes
6x + 5x − 6 = 0
2
... (i)
Splitting the middle term, we have
6x 2 + 9x – 4x − 6 = 0
or, 3x (2x + 3) –2 (2x + 3) = 0
or, (2x + 3)(3x – 2) = 0
3
∴ Either 2x + 3 = 0 ⇒ x = −
2
2
or, 3x – 2 = 0 ⇒ x =
3
3 2
∴ Two roots of the given quadratic equation are − ,
2 3
Example 2.2 Using factorization method, solve the following quadratic equation:
3 2 x + 7x − 3 2 = 0
2
Solution: Splitting the middle term, we have
3 2 x + 9x – 2x − 3 2 = 0
2
or, 3x ( 2 x + 3) − 2 ( 2 x + 3) = 0
or, ( 2 x + 3)(3x − 2 ) = 0
3
∴ Either 2x+3=0 ⇒ x= − 2
2
or, 3x − 2 = 0 ⇒ x=
3
3 2
∴ Two roots of the given quadratic equation are − ,
2 3
MATHEMATICS 41

4. Quadratic Equations
MODULE - I Example 2.3 Using factorization method, solve the following quadratic equation:
Algebra
(a + b)2 x2 + 6 (a2 − b2) x + 9 (a – b)2 = 0
Solution: The given quadratic equation is
(a + b)2 x2 + 6 (a2 − b2) x + 9 (a – b)2 = 0
Notes Splitting the middle term, we have
(a + b)2 x2 + 3(a2 − b2) x + 3(a2 – b2) x + 9 (a – b)2 = 0
or, (a + b) x {(a + b) x + 3 (a − b) } + 3 (a – b) {(a +b) x + 3 (a – b) } = 0
or, {(a + b) x + 3 (a − b) } {(a + b) x + 3 (a – b) } = 0
− 3(a − b) 3 (b − a)
∴ either (a + b) x + 3 (a – b) =0 ⇒ x = =
a+b a+b
− 3(a − b) 3 (b − a)
or, (a + b) x + 3 (a – b) =0 ⇒ x = =
a+b a+b
3 (b − a) 3 (b − a)
The equal roots of the given quadratic equation are ,
a+b a+b
Alternative Method
The given quadratic equation is
(a + b)2 x2 + 6(a2 − b2) x + 9(a – b)2 = 0
This can be rewritten as
{(a + b) x}2 + 2 .(a + b)x . 3 (a – b) + {3(a – b)}2 = 0
or, { (a + b)x + 3(a – b) }2 = 0
3(a − b) 3 (b − a)
or, x= − =
a+b a+b
3 (b − a) 3 (b − a)
∴ The quadratic equation have equal roots ,
a+b a+b
CHECK YOUR PROGRESS 2.1
1. Solve each of the following quadratic equations by factorization method:
2
(i) 3 x + 10x + 8 3 =0 (ii) x2 – 2ax + a2 – b = 0
 ab c 
(iii) x2 +  −  x – 1 = 0 (iv) x2 – 4 2x+6=0
 c ab 
42 MATHEMATICS

5. Quadratic Equations
Quadratic Equations
MODULE - I
2.3 SOLVING QUADRATIC EQUATION BY QUADRATIC
Algebra
FORMULA
Recall the solution of a standard quadratic equation
a x2 + bx + c = 0, a ≠ 0 by the “Method of Completing Squares”
Roots of the above quadratic equation are given by Notes
− b + b 2 − 4ac − b − b 2 − 4ac
x1= and x 2 =
2a 2a
−b+ D −b− D
= , =
2a 2a
where D = b2 – 4ac is called the discriminant of the quadratic equation.
For a quadratic equation ax 2 + bx + c = 0, a ≠ 0 if
(i) D>0, the equation will have two real and unequal roots
(ii) D=0, the equation will have two real and equal roots and both roots are
b
equal to −
2a
(iii) D<0, the equation will have two conjugate complex (imaginary) roots.
Example 2.4 Examine the nature of roots in each of the following quadratic equations and
also verify them by formula.
(i) x2 + 9 x +10 = 0 (ii) 9 y2 − 6 2 y + 2 = 0
2 t − 3t + 3 2 = 0
2
(iii)
Solution:
(i) The given quadratic equation is x2 + 9 x + 10 = 0
Here, a = 1, b = 9 and c= 10
∴ D = b2 – 4ac = 81 – 4.1.10
= 41>0.
∴ The equation will have two real and unequal roots
MATHEMATICS 43

6. Quadratic Equations
MODULE - I
Verification: By quadratic formula, we have
Algebra
− 9 ± 41
x=
2
− 9 + 41 − 9 − 41
Notes ∴ The two roots are , which are real and unequal.
2 2
(ii) The given quadratic equation is 9 y2 – 6 2 y + 2 = 0
Here, D = b2 – 4 ac
= ( − 6 2 )2 – 4.9.2
= 72 – 72 = 0
∴ The equation will have two real and equal roots.
Verification: By quadratic formula, we have
6 2± 0 2
y= =
2.9 3
2 2
∴ The two equal roots are , .
3 3
2 t − 3t + 3 2 = 0
2
(iii) The given quadratic equation is
Here, D = ( − 3)2 − 4. 2 .3 2
= − 15< 0
∴ The equation will have two conjugate complex roots.
Verification: By quadratic formula, we have
3 ± −15
t=
2 2
3 ± 15i
= , where i = − 1
2 2
3+ 15 i 3− 15 i
∴ Two conjugate complex roots are ,
2 2 2 2
44 MATHEMATICS

7. Quadratic Equations
Quadratic Equations
Example 2.5 Prove that the quadratic equation x2 + py – 1 = 0 has real and distinct roots
MODULE - I
for all real values of p. Algebra
Solution: Here, D = p2 + 4 which is always positive for all real values of p.
∴ The quadratic equation will have real and distinct roots for all real values of p.
Example 2.6 For what value of k the quadratice equation Notes
(4k+ 1) x2 + (k + 1) x + 1 = 0 will have equal roots ?
Solution: The given quadratic equation is
(4k + 1)x2 + (k + 1) x + 1 = 0
Here, D = (k + 1)2 – 4.(4k + 1).1
For equal roots, D = 0
∴ (k + 1)2 – 4 (4k + 1) = 0
⇒ k 2 – 14k –3 = 0
14 ± 196 +12
∴ k=
2
14 ± 208
or k=
2
= 7 ± 2 13 or 7 + 2 13 , 7 – 2 13
which are the required values of k.
Example 2.7 Prove that the roots of the equation
x2 (a2+ b2) + 2x (ac+ bd) + (c2+ d2) = 0 are imaginary. But if ad = bc,
roots are real and equal.
Solution: The given equation is x2 (a2 + b2) + 2x (ac + bd) + (c2 + d2) = 0
Discriminant = 4 (ac + bd)2 – 4 (a2 + b2) (c2 + d2)
= 8 abcd – 4(a2 d2 + b2 c2)
= − 4 ( − 2 abcd + a2 d2 + b2 c2)
= − 4 (ad –bc) 2
<0 for all a, b, c, d
∴ The roots of the given equation are imaginary.
For real and equal roots, discriminant is equal to zero.
⇒ − 4 (ad − bc)2 = 0
or, ad = bc
Hence, if ad=bc, the roots are real and equal.
MATHEMATICS 45

8. Quadratic Equations
MODULE - I
Algebra
CHECK YOUR PROGRESS 2.2
1. Solve each of the following quadratic equation by quadratic formula:
Notes (i) 2x2 – 3 x + 3 = 0 (ii) − x 2 + 2x − 1 = 0
(iii) −4 x 2 + 5 x − 3 = 0 (iv) 3x 2 + 2 x + 5 = 0
2. For what values of k will the equation
y2 – 2 (1 + 2k) y + 3 + 2k = 0 have equal roots ?
3. Show that the roots of the equation
(x − a) (x − b) + (x − b) (x − c) + (x − c) (x − a) = 0 are always real and they can not be
equal unless a = b = c.
2.4 RELATION BETWEEN ROOTS AND COEFFICIENTS OF
A QUADRATIC EQUATION
You have learnt that, the roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0
− b + b 2 − 4 ac − b − b 2 − 4 ac
are and
2a 2a
− b + b 2 − 4 ac − b − b 2 − 4 ac
Let  = ...(i) and  = ... (ii)
2a 2a
Adding (i) and (ii), we have
− 2b −b
 + = =
2a a
coefficient of x b
∴ Sum of the roots = – coefficient of x 2 = − a ... (iii)
+ b 2 − (b 2 − 4 ac)
  =
4a 2
4 ac
=
4a 2
c
=
a
46 MATHEMATICS

9. Quadratic Equations
Quadratic Equations
MODULE - I
∴
constant term c
Product of the roots = coefficient of x 2 = a ...(iv) Algebra
(iii) and (iv) are the required relationships between roots and coefficients of a given quadratic
equation. These relationships helps to find out a quadratic equation when two roots are
given.
Notes
Example 2.8 If,  ,  are the roots of the equation 3x2 – 5x + 9 = 0 find the value of:
1 1
(a)  2 +  2 (b) + 2
2
Solution: (a) It is given that  ,  are the roots of the quadratic equation 3x2 – 5x +9 = 0.
5
∴  + = ... (i)
3
9
and  = =3 ... (ii)
3
2 + 2 = (  +  ) − 2
2
Now,
2
 5
=   − 2.3 [By (i) and (ii)]
 3
29
=−
9
29
∴ The required value is −
9
1 1 2 +2
(b) Now, + 2 =
2  2 2
− 29
= 9 [By (i) and (ii)]
9
29
=−
81
MATHEMATICS 47

10. Quadratic Equations
MODULE - I Example 2.9 If  ,  are the roots of the equation 3y2 + 4y + 1 = 0, form a quadratic
Algebra
equation whose roots are  2,  2
Solution: It is given that  ,  are two roots of the quadratic equation 3y2 + 4y + 1 = 0.
∴ Sum of the roots
Notes
coefficient of y
i.e.,  +  = − coefficient of y 2
4
=− ... (i)
3
cons tan t term
Product of the roots i.e.,   = coefficient of y 2
1
= ... (ii)
3
Now,  2 +  2 = (  +  )2 – 2  
2
 4 1
=  −  − 2. [ By (i) and (ii)]
 3 3
16 2
= −
9 3
10
=
9
1
and  2  2 = (   )2 = [By (i) ]
9
∴ The required quadratic equation is y2 – (  2 +  2)y +  2  2 = 0
10 1
or, y2 − y+ =0
9 9
or, 9y2 – 10y + 1 = 0
Example 2.10 If one root of the equation ax 2 + bx + c = 0, a ≠ 0 be the square of
the other, prove that b3 + ac2 + a2c = 3abc
Solution: Let  ,  2 be two roots of the equation ax2 + bx + c = 0.
b
∴  + 2= − ... (i)
a
48 MATHEMATICS

11. Quadratic Equations
Quadratic Equations
MODULE - I
and  . 2 =
c Algebra
a
c
i.e., 3= . ... (ii)
a
Notes
From (i) we have
b
 (  + 1) = −
a
3
 b b3
or, { ( + 1)} 3
= −  = − 3
 a a
b3
or,  3 (  3 + 3  2 + 3  +1) = −
a3
c c  b  3
or,  + 3  −  +1 = − b3 ... [ By (i) and (ii)]
a a  a  a
c2 3 bc c b3
or, – + =– 3
a2 a2 a a
or, ac2 – 3abc + a2c = − b3
or, b3 + ac2 + a2c = 3abc
which is the required result.
Example 2.11 Find the condition that the roots of the equation ax2 + bx + c = 0 are in the
ratio m : n
Solution: Let m  and n  be the roots of the equation ax2 +bx + c = 0
b
Now, m  + n  = – ... (i)
a
c
and mn  2 = ... (ii)
a
b
From (i) we have,  (m + n ) = –
a
b2
or,  (m + n)2 =
2
a2
MATHEMATICS 49

12. Quadratic Equations
MODULE - I c b2
Algebra or,
a
(m + n)2 = mn 2
a
[ By (ii)]
or, ac (m+n)2 = mn b2
which is the required condition
Notes
CHECK YOUR PROGRESS 2.3
1. If  ,  are the roots of the equation ay2 + by + c = 0 then find the value of :
1 1 1 1
(i) + 2 (ii) + 4
 2
4
2. If  ,  are the roots of the equation 5x2 – 6x + 3 = 0, form a quadratic equation
whose roots are:
(i)  2,  2 (ii)  3  ,   3
3. If the roots of the equation ay2 + by + c = 0 be in the ratio 3:4, prove that
12b2 = 49 ac
4. Find the condition that one root of the quadratic equation px2 – qx + p = 0
may be 1 more than the other.
2.5 SOLUTION OF A QUADRATIC EQUATION WHEN D < 0
Let us consider the following quadratic equation:
(a) Solve for t :
t2 + 3t + 4 = 0
− 3 ± 9 −16 −3 ± −7
∴ t= =
2 2
Here, D= –7 < 0
−3 + −7 −3− − 7
∴ The roots are and
2 2
− 3 + 7 i −3 − 7 i
or, ,
2 2
Thus, the roots are complex and conjugate.
50 MATHEMATICS

13. Quadratic Equations
Quadratic Equations
(b) Solve for y : MODULE - I
Algebra
− 3y2 + 5 y – 2 = 0
− 5 ± 5 − 4(−3).(−2)
∴ y=
2(−3)
Notes
− 5 ± −19
or y=
−6
Here, D = − 19 < 0
− 5 + 19i 5 − 19i
∴ The roots are ,−
−6 −6
Here, also roots are complex and conjugate. From the above examples , we can make the
following conclusions:
(i) D < 0 in both the cases
(ii) Roots are complex and conjugate to each other.
Is it always true that complex roots occur in conjugate pairs ?
Let us form a quadratic equation whose roots are
2 + 3i and 4 – 5i
The equation will be {x – (2 + 3i)} {x – (4 – 5i)} = 0
or, x2 – (2 + 3i)x – (4 – 5i)x + (2 + 3i) (4 – 5i) = 0
or, x2 + (–6 + 2i)x + 23 + 2i = 0
which is an equation with complex coefficients.
Note : If the quadratic equation has two complex roots, which are not conjugate
of each other, the quadratic equation is an equation with complex coefficients.
2.6 CUBE ROOTS OF UNITY
Let us consider an equation of degree 3 or more. Any equation of degree 3 can be expressed
as a product of a linear and quadratic equation.
The simplest situation that comes for our consideration is
x³ − 1 = 0 ...(i)
∴ x³ − 1 = (x − 1) (x ² + x + 1) = 0
either x − 1 = 0 or, x²+x+1=0
MATHEMATICS 51

14. Quadratic Equations
MODULE - I
Algebra −1± 3i
or, x=1 or, x=
2
1 3i 1 3i
∴ Roots are 1, − + and − −
2 2 2 2
Notes
These are called cube roots of unity.
Do you see any relationship between two non-real roots of unity obtained above ?
Let us try to find the relationship between them
1 3i
Let w= − +
2 2
Squaring both sides, we have
2
 1 3 i
= − +
 2 
2 
w²
 
1
= (1+ 3 i 2 − 2 3 i)
4
1
= (1− 3 − 2 3 i)
4
−2−2 3 i
=
4
=
(
− 2 1+ 3 i ) =
(
− 1+ 3 i )
4 2
1 3i
∴ w² = − − = other complex root.
2 2
∴ We denote cube roots of unity as 1, w, w²
52 MATHEMATICS

15. Quadratic Equations
Quadratic Equations
MODULE - I
1
Note: If we take w = − −
3i
, we can verify, that in this case Algebra
2 2
1 3i
also w2 = − + = other complex root.
2 2
Notes
Thus, square of one complex root is same as the other complex root.
Again, sum of roots i.e., 1+ w + w²
 1 3 i  1 3 i
1 + − +
 2  + − − 
2   2 2 
=
   
1 1
= 1− −
2 2
= 1−1
= 0
∴ 1 + w + w² = 0
Sum of cube roots of unity is zero.
Product of cube roots of 1
i.e., 1.w.w² = w³ = 1 (since ,w is a root the equation x3 − 1 = 0)
We can conclude that , if 1, w and w2 are cube roots of unity then
(i) square of one complex root is same as the other complex root i.e., w2 = w
(ii) 1+ w + w2 = 0
(iii) w3 = 1
Let us now consider another equation
x³ + 1 = 0 ... (ii)
or (x + 1) (x ² − x + 1) = 0
Either, x + 1 = 0 or, x² −x+1=0
1± 1− 4
or, x = −1 or, x=
2
1± −3
⇒ x = −1 or, x=
2
MATHEMATICS 53

16. Quadratic Equations
MODULE - I
Algebra ∴ Roots are − 1,
1
−
3i 1
and +
3i
which can also be written as − 1 , − w and − w2
2 2 2 2
Therefore, cube roots of − 1 are − 1 , − w and − w2
In general, roots of any cubic equation of the form x 3 = ± a3 would be
Notes
± a , ± aw and ± aw2
Example 2.12 If 1, w and w² are cube roots of unity, prove that
(a) 1 + w2 + w7 = 0
(b) (1 − w + w2) (1 + w – w2) = 4
(c) (1 + w)3 − (1 + w2)3 = 0
(d) (1 − w + w2)3 = − 8 and (1 + w − w2)3 = − 8
Solution:
(a) 1 + w2 + w7 = 0
L.H.S = 1 + w2 + (w3)2. w
= 1 + w2 + w [ since w3 =1]
=0 [ since 1+ w + w2 = 0]
= R.H.S
∴ L.H.S = R.H.S
(b) (1 − w + w2) (1 + w – w2) = 4
L.H.S = (1 − w + w2) (1 + w – w2)
= (1 + w2 − w) (1 + w – w2)
(since 1 + w2 = − w and 1 + w = − w2)
= ( − w − w) ( − w2 –w2)
= ( − 2w) ( − 2w2)
= 4w3
= 4.1 = 4 =R.H.S
∴ L.H.S = R.H.S
54 MATHEMATICS

17. Quadratic Equations
Quadratic Equations
(c) (1 + w)3 − (1 + w2)3 = 0
MODULE - I
Algebra
L.H.S = (1 + w)3 − (1 + w2)3
= ( − w2)3 − ( − w)3 (∴ 1 + w = − w2
= − w6 + w3 and 1 + w2 = − w)
= − (w3)2 + 1 Notes
= − (1)2 + 1
= 0 =R.H.S
∴ L.H.S = R.H.S
(d) (1 − w + w2)3 = − 8 and (1 + w − w2)3 = − 8
Case I : L.H.S = (1 − w + w2)3
= (1 + w2 − w)3
= ( − w − w)3
= ( − 2w)3
= − 8w3
= − 8 = R.H.S
∴ L.H.S = R.H.S
Case II : L.H.S = (1 + w − w²)³
= ( − w2 – w2)3
= ( − 2w2)3
= − 8w6
= − 8(w3)2
= − 8 = R.H.S
∴ L.H.S = R.H.S
CHECK YOUR PROGRESS 2.4
1. Solve each of the following cubic equations:
(i) x3 = 27 (ii) x3 = − 27
(iii) x3 = 64 (iv) x3 = − 64
MATHEMATICS 55

18. Quadratic Equations
MODULE - I
2. If 1, w, w2 are cube roots of unity, show that
Algebra
(i) (1 + w) (1 + w2) (1 + w4) (1 + w8) = 1
(ii) (1 − w) (1 − w2) (1 – w4) (1 – w5) = 9
Notes (iii) (1 + w)4 + (1 + w2)4 = − 1
(iv) (1 + w3)3 = 8
(v) (1 – w + w2)6 = (1 + w – w2)6 = 64
(vi) (1 + w)16 + w = (1 + w2)16 + w2 = − 1
LET US SUM UP
• Roots of the quadratic equation ax2 + bx + c = 0 are complex and conjugate of each
other, when D < 0.
• If  ,  be the roots of the quadratic equation
b c
ax2 + bx + c = 0 then  +  = − and   =
a a
• Cube roots of unity are 1, w, w2
1 3i 1 3i
where w = − + and w 2 = − −
2 2 2 2
• Sum of cube roots of unity is zero i.e., 1+ w + w2 =0
• Product of cube roots of unity is 1 i.e., w3 = 1
• Complex roots w and w2 are conjugate to each other.
• In general roots of any cubic equation of the form
x 3 = ± a3 would be ± a , ± aw and ± aw2
56 MATHEMATICS

19. Quadratic Equations
Quadratic Equations
MODULE - I
Algebra
SUPPORTIVE WEB SITES
http://www.wikipedia.org
http://mathworld.wolfram.com
Notes
TERMINAL EXERCISE
1. Show that the roots of the equation
2(a2 + b2 )x2 + 2(a + b)x + 1=0 are imaginary,when a ≠ b
2. Show that the roots of the equation
bx2 + ( b – c)x = c + a –b are always real if those of
ax2 + b( 2x + 1) = 0 are imaginary.
3. If  ,  be the roots of the equation 2x2 – 6x + 5 = 0 , find the equation whose
roots are:
  1 1 1 1
(i)  , (ii)  +  ,  + (iii)  2 +  2 , + 2
  2
4. If 1, w and w2 are cube roots of unity , prove that
(a) (2 – w) (2 – w2) (2 – w10) (2 – w11) = 49
(b) ( x – y) (xw – y) ) (xw2 – y) = x3 – y3
5. If x = a + b , y = aw + bw2 and z = aw2 + bw , then prove that
(a) x2 + y2 + z2 = 6 ab (b) x y z = a3 + b3
MATHEMATICS 57

20. Quadratic Equations
MODULE - I
Algebra
ANSWERS
CHECK YOUR PROGRESS 2.1
Notes
−4
1. (i) −2 3 , (ii) a − b, a + b
3
ab c
(iii) − , (iv) 3 2, 2
c ab
CHECK YOUR PROGRESS 2.2
3 ± 15 i 1± i
1. (i) (ii)
4 2
5 ± 43 i − 2 ± 58 i
(iii) (iv)
8 6
1
2. –1,
2
CHECK YOUR PROGRESS 2.3
b 2 − 2ac (b 2 − 2ac)2 − 2a 2 c 2
1. (i) (ii)
c2 c4
2. (i) 25x2 – 6x + 9 = 0 (ii) 625x2 – 90x + 81 = 0
4. q2 – 5p2 = 0
CHECK YOUR PROGRESS 2.4
1. (i) 3, 3w, 3w2 (ii) − 3, − 3w , − 3w2
(iii) 4, 4w, 4w2 (iv) − 4, − 4w , − 4w2
TERMINAL EXERCISE
3. (i) 5x2 – 8x + 5 = 0 (ii) 10x2 – 42x + 49 = 0 (iii) 25x2 – 116x + 64 = 0
58 MATHEMATICS